Question

In Problem, find the area of the parallelogram with one corner at P1 and adjacent sides

$\overrightarrow{P1P2}and\overrightarrow{P1P3}$

$P1=\left(0,0,0\right),P2=\left(2,3,1,\right)P3=\left(-2,4,1\right)$

Short answer

The area of the parallelogram is $\sqrt{213}$

Step-by-step solution

Step1. Given equation 

three points$P1=\left(0,0,0\right),P2=\left(2,3,1\right),P3=\left(-2,4,1\right)$

We know that the vector is formed by two points like

To given question

$$\begin{gathered} U=\overrightarrow{P1P2} \\ U=\left(2-0\right)i+\left(3-0\right)j+\left(1-0\right)k \\ U=2i+3j+k \\ likethatv=-2i+4j+k \end{gathered}$$

Step2. Calculation of cross product 

We know that to find the area of a parallelogram by two vectors. it's necessary to find an orthogonal vector, for that

$$\begin{gathered} u\times v=\left|\begin{array}{ccc}i& j& k\\ 2& 3& 1\\ -2& 4& 1\end{array}\right| \\ u\times v=\left[i\left|\begin{array}{cc}3& 2\\ 4& 2\end{array}\right|+j\left|\begin{array}{cc}2& 1\\ -2& 1\end{array}\right|+k\left|\begin{array}{cc}2& 3\\ -2& 4\end{array}\right|\right] \\ u\times v=-i-4j+14k \end{gathered}$$

Step3.Calculation of area 

The area of parallelogram = $\left|\left|u\times v\right|\right|$

$$\begin{gathered} \left|\left|u\times v\right|\right|=\left|-i-4j+14k\right| \\ \left|\left|u\times v\right|\right|=\sqrt{1+16+196} \\ \left|\left|u\times v\right|\right|=\sqrt{213} \end{gathered}$$